TITLE: Putting objects in a line.
QUESTION [6 upvotes]: I'm working on a project outside of school, and I've run into a bit a problem.  I thought, maybe there are some problem solvers on the internet who would enjoy this.
I have 8 balls, 3 red cubes, and 4 blue cubes.  
How many ways can I arrange these items in a line so that:

I never have three cubes consecutively
I never have three balls consecutively
I never have two red cubes next to one another
I never have two blue cubes next to one another

If it's really hard or if it doesn't seem like an interesting problem, no worries.  Basically, I want to understand exactly what these restrictions limit me to.
If I come up with anything myself, I'll post it.

REPLY [4 votes]: Here's one more to supplement Rebecca's answer:
We may use a directed graph for such problems, and for our problem, the adjacency matrix of such graph is:
\begin{align*}
 \begin{array}{|l|rrrrrr|}\hline
 & \mathrm{I} & \mathrm{R} & \mathrm{B} & \mathrm{C} & \mathrm{BR} & \mathrm{CC} \\ \hline
\mathrm{I} & 0 & r & b & c & 0 & 0 \\
\mathrm{R} & 0 & 0 & 0 & c & b & 0 \\
\mathrm{B} & 0 & 0 & 0 & c & r & 0 \\
\mathrm{C} & 0 & r & b & 0 & 0 & c \\
\mathrm{BR} & 0 & 0 & 0 & c & 0 & 0 \\
\mathrm{CC} & 0 & r & b & 0 & 0 & 0 \\ \hline
\end{array}
\end{align*}
where R indicates red cubes, B indicates blue cubes and C is for balls.
Since there are 15 objects, we compute
\begin{align*}
  \left(\begin{array}{rrrrrr}
0 & r & b & c & 0 & 0 \\
0 & 0 & 0 & c & b & 0 \\
0 & 0 & 0 & c & r & 0 \\
0 & r & b & 0 & 0 & c \\
0 & 0 & 0 & c & 0 & 0 \\
0 & r & b & 0 & 0 & 0
\end{array}\right)^{15}
\end{align*}
and take the sum of the first row, and we see that $[c^8b^4r^3]=\boxed{11394}$ 
It's also possible to get the following multivariate generating function:
\begin{align*}
  G(r,b,c) &= \frac{\left(1+c+c^2\right)\left(1+b+r+2br\right)}{1-c\left(1+c\right)\left(b+r+2br\right)}
\end{align*}
Update
If we compute the characteristic polynomial of the matrix, we can see how the recurrence relation is structured:
\begin{align*}
  x^6  -(bc + cr)x^4 - (bc^2 + 2bcr + c^2r)x^3 - 2bc^2rx^2 = 0
\end{align*}
and the required recurrence is:
\begin{align*}
  f_{b,c,r} &= f_{b-1,c-1,r}+f_{b,c-1,r-1}+f_{b-1,c-2,r}+f_{b,c-2,r-1}+2 \left(f_{b-1,c-1,r-1}+f_{b-1,c-2,r-1}\right)
\end{align*}
and set the boundary conditions like $f_{0,0,0}=1, f_{1,0,0}=f_{0,1,0}=f_{0,0,1}=1$ etc.
We find that $f_{4,8,3}=11394$
With some perseverance, I think it's also possible to get a summation from the generating function.